2009 AIME I Problems/Problem 6
Contents
Problem
How many positive integers less than
are there such that the equation
has a solution for
?
Solution
First, must be less than
, since otherwise
would be at least
which is greater than
.
Because must be an integer, we can do some simple case work:
For ,
as long as
. This gives us
value of
.
For ,
can be anything between
to
excluding
Therefore, . However, we got
in case 1 so it got counted twice.
For ,
can be anything between
to
excluding
This gives us
's
For ,
can be anything between
to
excluding
This gives us
's
For ,
can be anything between
to
excluding
This gives us
's
Since must be less than
, we can stop here and the answer is
possible values for
.
Alternatively, one could find that the values which work are to get the same answer.
Solution 2
For a positive integer , we find the number of positive integers
such that
has a solution with
. Then
, and because
, we have
, and because
is an integer, we get
. The number of possible values of
is equal to the number of integers between
and
, which is equal to the larger number minus the smaller number plus one or
, and this is equal to
. If
, the value of
exceeds
, so we only need to consider
. The requested number of values of
is the same as the number of values of
, which is
.
Video Solution
Mostly the above solution explained on video: https://www.youtube.com/watch?v=2Xzjh6ae0MU&t=11s
~IceMatrix
Video Solution 2
~Shreyas S
Video Solution 3
Projective Solution: https://youtu.be/fUef_tVnM5M
~Shreyas S
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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